http://jipam.vu.edu.au/
Volume 6, Issue 2, Article 39, 2005
ON A NEW MULTIPLE EXTENSION OF HILBERT’S INTEGRAL INEQUALITY
BICHENG YANG DEPARTMENT OFMATHEMATICS, GUANGDONGINSTITUTE OFEDUCATION,
GUANGZHOU, GUANGDONG510303, PEOPLE’SREPUBLICOFCHINA. bcyang@pub.guangzhou.gd.cn
URL:http://page.gdei.edu.cn/ yangbicheng
Received 04 June, 2004; accepted 18 March, 2005 Communicated by B.G. Pachpatte
ABSTRACT. This paper gives a new multiple extension of Hilbert’s integral inequality with a best constant factor, by introducing a parameterλand theΓfunction. Some particular results are obtained.
Key words and phrases: Hilbert’s integral inequality; Weight coefficient ,Γfunction.
2000 Mathematics Subject Classification. 26D15.
1. INTRODUCTION
Iff, g ≥0satisfy 0<
Z ∞ 0
f2(x)dx <∞ and 0<
Z ∞ 0
g2(x)dx <∞,
then (1.1)
Z ∞ 0
Z ∞ 0
f(x)g(y)
x+y dxdy < π Z ∞
0
f2(x)dx Z ∞
0
g2(x)dx 12
,
where the constant factor π is the best possible (cf. Hardy et al. [2]). Inequality (1.1) is well known as Hilbert’s integral inequality, which had been extended by Hardy [1] as:
Ifp > 1, 1p +1q = 1, f, g ≥0satisfy 0<
Z ∞ 0
fp(x)dx <∞ and 0<
Z ∞ 0
gq(x)dx <∞,
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
144-04
~
then (1.2)
Z ∞ 0
Z ∞ 0
f(x)g(y)
x+y dxdy < π sin
π p
Z ∞
0
fp(x)dx
1pZ ∞ 0
gq(x)dx 1q
,
where the constant factor sin(π/p)π is the best possible. Inequality (1.2) is called Hardy- Hilbert’s integral inequality, and is important in analysis and its applications (cf. Mitrinovi´c et al.[6]).
Recently, by introducing a parameterλ,Yang [9] gave an extension of (1.2) as:
Ifλ >2−min{p, q}, f, g ≥0satisfy 0<
Z ∞ 0
x1−λfp(x)dx <∞ and 0<
Z ∞ 0
x1−λgq(x)dx <∞,
then (1.3)
Z ∞ 0
Z ∞ 0
f(x)g(y)
(x+y)λdxdy < kλ(p) Z ∞
0
x1−λfp(x)dx
1pZ ∞ 0
x1−λgq(x)dx 1q
, where the constant factor kλ(p) = Bp+λ−2
p ,q+λ−2q
is the best possible (B(u, v) is the β function). Forλ= 1,inequality (1.3) reduces to (1.2).
On the problem for multiple extension of (1.1), [3, 4] gave some new results and Yang [8]
gave an improvement of their works as:
Ifn ∈N\{1}, pi >1,Pn i=1
1
pi = 1, λ > n− min
1≤i≤n{pi}, fi ≥0,satisfy 0<
Z ∞ 0
xn−1−λfipi(x)dx <∞ (i= 1,2, . . . , n),
then (1.4)
Z ∞ 0
· · · Z ∞
0
1 Pn
j=1xjλ n
Y
i=1
fi(xi)dx1. . . dxn
< 1 Γ(λ)
n
Y
i=1
Γ
pi+λ−n pi
Z ∞ 0
xn−1−λfipi(x)dx pi1
,
where the constant factorΓ(λ)1 Qn i=1Γ
pi+λ−n pi
is the best possible. Forn= 2, inequality (1.4) reduces to (1.3). It follows that (1.4) is a multiple extension of (1.3), (1.2) and (1.1).
In 2003, Yang et. al [11] provided an extensive account of the above results.
The main objective of this paper is to build a new extension of (1.1) with a best constant factor other than (1.4), and give some new particular results. That is
Theorem 1.1. Ifn∈N\{1}, pi >1,Pn i=1
1
pi = 1, λ >0, fi ≥0,satisfy 0<
Z ∞ 0
xpi−1−λfipi(x)dx <∞ (i= 1,2, . . . , n),
then (1.5)
Z ∞ 0
· · · Z ∞
0
1 Pn
j=1xjλ n
Y
i=1
fi(xi)dx1. . . dxn
< 1 Γ(λ)
n
Y
i=1
Γ λ
pi
Z ∞ 0
xpi−1−λfipi(x)dx pi1
,
where the constant factor Γ(λ)1 Qn i=1Γ
λ pi
is the best possible. In particular,
(a) forλ= 1,we have (1.6)
Z ∞ 0
· · · Z ∞
0
Qn
i=1fi(xi) Pn
j=1xj
dx1. . . dxn <
n
Y
i=1
Γ 1
pi
Z ∞ 0
xpi−2fipi(x)dx pi1
;
(b) forn= 2, using the symbol of (1.3) and settingekλ(p) = B
λ p,λq
,we have
(1.7) Z ∞
0
Z ∞ 0
f(x)g(y)
(x+y)λdxdy <ekλ(p) Z ∞
0
xp−1−λfp(x)dx
1pZ ∞ 0
xq−1−λgq(x)dx 1q
,
where the constant factors in (1.6) and (1.7) are still the best possible.
In order to prove the theorem, we introduce some lemmas.
2. SOME LEMMAS
Lemma 2.1. Ifk ∈N, ri >1 (i= 1,2, . . . , k+ 1),andPk+1
i=1 ri =λ(k),then
(2.1)
Z ∞ 0
· · · Z ∞
0
1
1 +Pk
j=1ujλ(k) k
Y
j=1
urjj−1du1. . . duk = Qk+1
i=1 Γ(ri) Γ(λ(k)) .
Proof. We establish (2.1) by mathematical induction. Fork = 1, sincer1+r2 =λ(1),and (see [7])
(2.2) B(p, q) =
Z ∞ 0
up−1
(1 +u)p+qdu=B(q, p) (p, q >0),
we have (2.1). Suppose for k ∈ N,that (2.1) is valid. Then fork + 1, sincer1 +Pk+1 i=2 ri = λ(k+ 1), by settingv =u1.
1 +Pk+1 j=2uj
, we obtain Z ∞
0
· · · Z ∞
0
1
1 +Pk+1
j=1ujλ(k+1) k+1
Y
j=1
urjj−1du1. . . duk+1 (2.3)
= Z ∞
0
· · · Z ∞
0
vr1−1
1 +Pk+1 j=2ujr1
Qk+1 j=2urjj−1
1 +Pk+1
j=2ujλ(k+1)
(1 +v)λ(k+1)
dvdu2. . . duk+1
= Z ∞
0
· · · Z ∞
0
Qk+1 j=2urjj−1
1 +Pk+1
j=2 ujλ(k+1)−r1du2. . . duk+1 Z ∞
0
vr1−1
(1 +v)λ(k+1)dv.
In view of (2.2) and the assumption ofk, we have (2.4)
Z ∞ 0
vr1−1
(1 +v)λ(k+1)dv = 1
Γ(λ(k+ 1))Γ
k+1
X
i=2
ri
! Γ(r1);
(2.5)
Z ∞ 0
· · · Z ∞
0
Qk+1 j=2urjj−1
1 +Pk+1
j=2ujλ(k+1)−r1du2. . . duk+1 =
Qk+2 i=2 Γ(ri) Γ
Pk+1 i=2 ri.
Then, by (2.5), (2.4) and (2.3), we find Z ∞
0
· · · Z ∞
0
Qk+1 j=1urjj−1
1 +Pk+1
j=1ujλ(k+1)du1. . . duk+1 = Qk+2
i=1 Γ(ri) Γ(λ(k+ 1)).
Hence (2.1) is valid fork ∈Nby induction. The lemma is proved.
Lemma 2.2. Ifn ∈ N\{1}, pi > 1 (i = 1,2, . . . , n),Pn i=1
1
pi = 1andλ > 0, set the weight coefficientω(xi)as
(2.6) ω(xi) :=x
λ pj
i
Z ∞ 0
· · · Z ∞
0
Qn
j=1(j6=i)x(λ−pj j)/pj Pn
j=1xjλ dx1. . . dxi−1dxi+1. . . dxn. Then, eachω(xi)is constant, that is
(2.7) ω(xi) = 1
Γ(λ)
n
Y
j=1
Γ λ
pj
, (i= 1,2, . . . , n).
Proof. Fixi. Settingpen =pi,andpej =pj, uj =xj/xi, forj = 1,2, . . . , i−1;pej =pj+1, uj = xj+1/xi, forj =i, i+ 1, . . . , n−1in (2.6), by simplification, we have
(2.8) ω(xi) = Z ∞
0
· · · Z ∞
0
1
1 +Pn−1 j=1 ujλ
n−1
Y
j=1
u
−1+λ
pj
j du1. . . dun−1.
Substitution ofn−1fork, λforλ(k)andλ/pej forrj (j = 1,2, . . . , n)into (2.1), in view of (2.8), we have
ω(xi) = 1 Γ(λ)
n
Y
j=1
Γ λ
pej
= 1
Γ(λ)
n
Y
j=1
Γ λ
pj
.
Hence, (2.7) is valid. The lemma is proved.
Lemma 2.3. As in the assumption of Lemma 2.2, for0< ε < λ, we have
I :=ε Z ∞
1
· · · Z ∞
1
Qn
i=1x(λ−pi i−ε)/pi Pn
j=1xjλ dx1. . . dxn
≥ 1
Γ(λ)
n
Y
i=1
Γ λ
pi
(ε→0+).
(2.9)
Proof. Settingui =xi/xn(i= 1,2, . . . , n−1)in the following, we find I =ε
Z ∞ 1
x−1−εn
Z ∞
x−1n
· · · Z ∞
x−1n
Qn−1
i=1 u(λ−pi i−ε)/pi
1 +Pn−1
j=1ujλdu1. . . dun−1
dxn
≥ε Z ∞
1
x−1−εn
Z ∞
0
· · · Z ∞
0
Qn−1
i=1 u(λ−pi i−ε)/pi
1 +Pn−1 j=1 uj
λdu1. . . dun−1
dxn
−ε Z ∞
1
x−1n
n−1
X
j=1
Aj(xn)dxn, (2.10)
where, forj = 1,2, . . . , n−1, Aj(xn)is defined by
(2.11) Aj(xn) :=
Z
· · · Z
Dj
Qn−1
i=1 u(λ−pi i−ε)/pi (1 +Pn−1
j=1 uj)λ du1. . . dun−1, satisfyingDj ={(u1, u2, . . . , un−1)|0< uj ≤x−1n , 0< uk<∞(k6=j)}.
Without loss of generality, we estimate the integralAj(xn)forj = 1.
(a) Forn = 2, we have A1(xn) =
Z x−1n
0
1
(1 +u1)λu(λ−p1 1−ε)/p1du1
≤ Z x−1n
0
u(λ−p1 1−ε)/p1du1 = p1
λ−εx−(λ−ε)/pn 1;
(b) Forn ∈N\{1,2},by (2.1), we have
A1(xn)≤ Z ∞
0
· · · Z ∞
0
Qn−1 i=2 u−1+
λ−ε pi
i
1 +Pn−1
j=2ujλdu1. . . dun−1
Z x−1n
0
u
λ−p1−ε p1
1 du1
≤ p1x−
λ−ε p1
n
λ−ε Z ∞
0
· · · Z ∞
0
Qn−1
i=2 u−1+(λ−ε)/pi i
1 +Pn−1 j=2 uj
(λ−ε)(1−p−11 )du1. . . dun−1
= p1x−
λ−ε p1
n
λ−ε ·
Qn
i=2Γ(λ−εp
i ) Γ((λ−ε)(1−p−11 )).
By virtue of the results of (a) and (b), forj = 1,2, . . . , n−1,we have (2.12) Aj(xn)≤ pj
λ−εx−(λ−ε)/pn jOj(1) (ε →0+, n∈N\{1}).
By (2.11), since forε →0+, Z ∞
0
· · · Z ∞
0
Qn−1
i=1 u−1+(λ−ε)/pi i
1 +Pn−1
j=1ujλ du1. . . dun−1 = Qn
i=1Γ
λ pi
Γ(λ) +o(1);
Z ∞ 1
x−1n
n−1
X
j=1
Aj(xn)dxn=
n−1
X
j=1
Z ∞ 1
x−1n Aj(xn)dxn
≤
n−1
X
j=1
pj
λ−εOj(1) Z ∞
1
x−1−(λ−ε)/pn jdxn
=
n−1
X
j=1
pj λ−ε
2
Oj(1),
then by (2.10) , we find I ≥
Qn
i=1Γ
λ pi
Γ(λ) +o(1)
−ε
n−1
X
j=1
pj λ−ε
2
Oj(1)
→ Qn
i=1Γ λ
pi
Γ(λ) (ε→0+).
Thereby, (2.9) is valid and the lemma is proved.
3. PROOF OF THE THEOREM ANDREMARKS
Proof of Theorem 1.1. By Hölder’s inequality, we have J :=
Z ∞ 0
· · · Z ∞
0
1 Pn
j=1xj
λ n
Y
i=1
fi(xi)dx1. . . dxn
= Z ∞
0
· · · Z ∞
0
n
Y
i=1
fi(xi) Pn
j=1xjλ/pi
x(pi−λ)(1−p
−1 i ) i
n
Y
j=1
(j6=i)
x
λ−pj pj
j
1 pi
dx1. . . dxn
≤
n
Y
i=1
Z ∞
0
· · · Z ∞
0
fipi(xi) Pn
j=1xjλx(pi−λ)(1−p
−1 i ) i
n
Y
j=1
(j6=i)
x
λ−pj pj
j dx1. . . dxn
1 pi
. (3.1)
If (3.1) takes the form of equality, then there exists constantsC1, C2, . . . , Cn,such that they are not all zero and for anyi6=k∈ {1,2, . . . , n}(see [5]),
Ci
fipi(xi)x(pi−λ)(1−p
−1 i ) i
Pn
j=1xjλ
n
Y
j=1
(j6=i)
x
λ−pj pj
j =Ck
fkpk(xk)x(pk−λ)(1−p
−1 k ) k
Pn
j=1xjλ
n
Y
j=1
(j6=k)
x
λ−pj pj
j ,
a.e. in(0,∞)× · · · ×(0,∞).
(3.2)
Assume thatCi 6= 0.By simplification of (3.2), we find
xpii−λfipi(xi) = F(x1, . . . , xi−1, xi+1, . . . , xn)
=constant a.e. in(0,∞)× · · · ×(0,∞), which contradicts the fact that0 < R∞
0 xpii−λ−1fipi(x)dx < ∞.Hence by (2.6) and (3.1), we conclude
(3.3) J <
n
Y
i=1
Z ∞ 0
ω(xi)xpii−1−λfipi(xi)dxi pi1
. Then by (2.7), we have (1.5).
For0< ε < λ,settingfei(xi)as: fei(xi) = 0,forxi ∈(0,1);
fei(xi) = x(λ−pi i−ε)/pi, forxi ∈[1,∞) (i= 1,2, . . . , n),
then we find
(3.4) ε
n
Y
i=1
Z ∞ 0
xpii−1−λfeipi(xi)dxi 1
pi = 1.
By (2.9), we find
(3.5) ε
Z ∞ 0
· · · Z ∞
0
1 Pn
j=1xjλ n
Y
i=1
fei(xi)dx1. . . dxn
=I ≥ 1 Γ(λ)
n
Y
i=1
Γ λ
pi
(ε→0+).
If the constant factorΓ(λ)1 Qn i=1Γ
λ pi
in (1.5) is not the best possible, then there exists a posi- tive constantK < Γ(λ)1 Qn
i=1Γ
λ pi
, such that (1.5) is still valid if one replacesΓ(λ)1 Qn i=1Γ
λ pi
byK.In particular, one has
ε Z ∞
0
· · · Z ∞
0
1 Pn
j=1xjλ n
Y
i=1
fei(xi)dx1. . . dxn< εK
n
Y
i=1
Z ∞ 0
xpi−1−λfeipi(x)dx 1
pi ,
and in view of (3.4) and (3.5), it follows that Γ(λ)1 Qn i=1Γ
λ pi
≤K(ε→0+).This contradicts the factK < Γ(λ)1 Qn
i=1Γ
λ pi
.Hence the constant factor Γ(λ)1 Qn i=1Γ
λ pi
in (1.5) is the best possible.
The theorem is proved.
Remark 3.1. Forλ= 1, inequality (1.7) reduces to (see [10])
(3.6)
Z ∞ 0
Z ∞ 0
f(x)g(y)
x+y dxdy < π sin
π p
Z ∞
0
xp−2fp(x)dx
p1 Z ∞ 0
xq−2gq(x)dx 1q
. Forp =q = 2, both (3.6) and (1.2) reduce to (1.1). It follows that inequalities (3.6) and (1.2) are different extensions of (1.1). Hence, inequality (1.5) is a new multiple extension of (1.1).
Since all the constant factors in the obtained inequalities are the best possible, we have obtained new results.
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